Applied Force Direction vector = [0,1,3]
force F = j + 4k
torque is t = 100 Newton-meters = r x F
Wrench is 0.5 meters long on positive side of the y-axis, r = 0.6 = [0,0.6,0]
We know torque |t| = |r x F| = |r| x |F| sin theta
r x F = |r| x |F| cos theta
r x (j + 3k) = |r| x |j + 3k| cos theta
=> [0,0.6,0] [0,1,3] = 0.6 x squareroot of ((0)^2 + (1)^2 + (3)^2) cos theta
=> 0.6 = 0.6 x squareroot of (1 + 9) cos theta
=> cos theta = 1 / squareroot of (10)
Calculationg the sin theta,
sin theta = squareroot of (1 - (1 / squareroot of (10))^2) = squareroot of (9/10)
sin theta = 3 / squareroot of (10)
Substituting the values,
|T| = |r| x |F| sin theta => 100 = |0.6| x |F| x 3 / squareroot of (10)
|F| = (100 x squareroot of (10)) / 1.8
|F| = (1000 / 18) x squareroot of (10)
Magnitude of force |F| = 55.55 x squareroot of (10)
